# Math Help - Splitting Fields - Dummit and Foote - Exercise 1, page 545

1. ## Splitting Fields - Dummit and Foote - Exercise 1, page 545

Dummit and Foote Exercise 1 on page 545 reads as follows:

Determine the splitting field and its degree over $\mathbb{Q}$ for $x^4 - 2$.

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I have started on the solution to this exercise as follows:

The two roots of $x^4 - 2$ are $\alpha = \sqrt[4]{2}$ and $\beta = \sqrt[4]{2}i$

Thus the splitting field is $\mathbb{Q}(\sqrt[4]{2}, \sqrt[4]{2}i )$

We note that $\mathbb{Q}(\sqrt[4]{2}, \sqrt[4]{2}i ) = \mathbb{Q}(\sqrt[4]{2}, i )$ since the product of $\sqrt[4]{2}$ and i must be in $\mathbb{Q}(\sqrt[4]{2}, i )$

The element $\alpha = \sqrt[4]{2} \in \mathbb{R}$ is algebraic over $\mathbb{Q}$ and so $m_{\alpha}(x) = m_{\sqrt[4]{2}}(x) = x^4 - 2$ is the minimal polynomial for $\alpha = \sqrt[4]{2}$ over $\mathbb{Q}$.

Thus the degree of $\alpha = \sqrt[4]{2}$ over $\mathbb{Q}$ is the degree of $m_{\alpha}(x)$ = 4

Hence $[\mathbb{Q}(\sqrt[4]{2}) \ : \ \mathbb{Q}]$ = 4

The element $\beta = \sqrt[4]{2}i \in \mathbb{C}$ is also algebraic over $\mathbb{Q}$ and so $m_{\beta}(x) = m_{\sqrt[4]{2}i}(x) = x^4 - 2$ is the minimal polynomial for $\beta = \sqrt[4]{2}i$ over $\mathbb{Q}$.

Thus the degree of $\beta = \sqrt[4]{2}i$ over $\mathbb{Q}$ is the degree of $m_{\beta}(x)$ = 4

Hence $[\mathbb{Q}(\sqrt[4]{2}i) \ : \ \mathbb{Q}]$ = 4

But where to from here - need to find the degree of the splitting field.

Can someone please confirm that my reasoning above is valid and show me the way forward from here?

Peter